Integrability in the product space can follow from a property of the Nemytskii operator?

Question

Let's say that $f:\Omega\times\mathbb{R}\to\mathbb{R}$ is a Caratheodory function (i.e. $f(x,\cdot)$ is continuous for a.a. $x\in\Omega$ and $f(\cdot,t)$ is measurable for all $t\in\mathbb{R}$), where $\Omega\subset\mathbb{R}^N$ is a bounded domain.

We assume that there are two reals $p,q>1$ such that for any $u\in L^{p}(\Omega)$ we have that the Nemytskii operator $\mathcal{N}_f(u):=f(\cdot,u(\cdot))\in L^{q}(\Omega)$.

Can we say that $f\in L^1_{\text{loc}}(\mathbb{R}; L^1(\Omega))$, i.e. for all $a<b$, $a,b\in\mathbb{R}$ we have that:

$$\int_{a}^{b}\int_{\Omega} |f(x,t)|\ dx\ dt<\infty$$

?

P.S. It is known that $f(x,t)$ is measurable in the product space $\Omega\times\mathbb{R}$, because it is a Caratheodory function.

Answer 1

Nemytskii operators on Lebesgue spaces are funny objects with a lot of implicit structure. In fact, if $\mathcal N_f$ maps $L^p(\Omega)$ into $L^q(\Omega)$, then $f$ must necessarily satisfy the growth condition $$\lvert f(x,u) \rvert \leq \gamma(x) + \beta \lvert u\rvert^{p/q}$$ with a $\gamma \in L^q(\Omega)$ and some $\beta \geq 0$. This even implies that $f \in L^q_{\text{loc}}(\mathbb R; L^q(\Omega))$ in your setting and notation if I am not mistaken.

I like the reference On Nemytskij Operators in $L_p$-Spaces of Abstract Functions by Goldberg et al. It even covers Nemytskii operators between Lebesgue-Bochner spaces.

(You will find also results regarding $p$ or $q = \infty$ there, I was not sure from your problem description whether that is also interesting for you.)